I posted a question about an I beam with a Moment of Inertia of 26.49.
It is a mild steel I-Beam and is 3.5" wide and 6" high. The beam
length is 38" and will have a 40,000 lb load placed in the middle of
the beam with a .5" spreader plate
where the jack will attach (This is going to be the top beam for a
hydraulic press). The deflection at load is .05". The beam will be
supported by two 60" posts that are 2.5"x2.5"x.25" thick square posts.
Someone states that this would "bend like a pretzel" under full load.
If the deflection is only .05" at maximum load how will the beam "bend
like a pretzel"? I know that the elastic yield strength of mild steel
is around 36,000 psi but I have read that most steel these days is
around the 45,000 psi. The same poster stated that the max stress will
be 43.6kpsi. I could truss the beam or could make the beam shorter in
length or could lower the hydraulic jack to a 10 ton model. I tried to
post this using the original post but was unsuccessful. Sorry for top
posting. Thanks, Steve.

Your project involves a lot more than just sizing a beam for the top of a
hydraulic press.
However, to answer your original question, for 40 kips, no impact sustained
load, at the center of a 3' span, A-36 steel, allowable f sub b =24,000psi,
you need a shape with a minimum section modulus of 15.

When the compression half of the cross-section of a steel beam is
overloaded, it acts like an overloaded column and can buckle sideways.
It won't bend like a pretzel, exactly, but it will fail........ that's what
he means.

"Steve" wrote in message
om...
I posted a question about an I beam with a Moment of Inertia of 26.49.
It is a mild steel I-Beam and is 3.5" wide and 6" high. The beam
length is 38" and will have a 40,000 lb load placed in the middle of
the beam with a .5" spreader plate
where the jack will attach (This is going to be the top beam for a
hydraulic press). The deflection at load is .05". The beam will be
supported by two 60" posts that are 2.5"x2.5"x.25" thick square posts.
Someone states that this would "bend like a pretzel" under full load.
If the deflection is only .05" at maximum load how will the beam "bend
like a pretzel"? I know that the elastic yield strength of mild steel
is around 36,000 psi but I have read that most steel these days is
around the 45,000 psi. The same poster stated that the max stress will
be 43.6kpsi. I could truss the beam or could make the beam shorter in
length or could lower the hydraulic jack to a 10 ton model. I tried to
post this using the original post but was unsuccessful. Sorry for top
posting. Thanks, Steve.

I posted a question about an I beam with a Moment of Inertia of 26.49.
It is a mild steel I-Beam and is 3.5" wide and 6" high. The beam
length is 38" and will have a 40,000 lb load placed in the middle of
the beam with a .5" spreader plate
where the jack will attach (This is going to be the top beam for a
hydraulic press). The deflection at load is .05". The beam will be
supported by two 60" posts that are 2.5"x2.5"x.25" thick square posts.
Someone states that this would "bend like a pretzel" under full load.
If the deflection is only .05" at maximum load how will the beam "bend
like a pretzel"? I know that the elastic yield strength of mild steel
is around 36,000 psi but I have read that most steel these days is
around the 45,000 psi. The same poster stated that the max stress will
be 43.6kpsi. I could truss the beam or could make the beam shorter in
length or could lower the hydraulic jack to a 10 ton model. I tried to
post this using the original post but was unsuccessful. Sorry for top
posting. Thanks, Steve.

I'm not a mechanical engineer but there are several steps to choosing a
beam. First, from the dimensions you calculate (or look up) the moment of
inertia, which with the modulus of elasticity gives the deflection under
some load. This you've done, and 0.05" is acceptable to you. Next you have
to calculate the shear stress in the beam under that load and compare the
stress to the material strength. In this case the material along the top of
the beam is going to be under tension and the material along the bottom of
the beam will be under compression as the beam flexes upwards from your jack
under the beam. Adjacent metal fibers will be under different amounts of
tension or compression and will want to slide past each other, or shear.
For this you need the section modulus which you also calculate or look up,
and from your post the shear stress turns out to be 43.6 ksi. I looked in
one table and found 36 ksi as the yield tension strength and 21 ksi as the
yield shear strength, plus you need a margin of safety on the order of four
times, so the shear stress has to be kept under 21 ksi/4=5.25 ksi (I said
I'm not an engineer; a real one would surely use a different safety margin
:-)). You are way over the acceptable stress limit of your beam, which
means that it's going to bend into a U before you get anywhere near your
40,000 lbs applied (not a pretzel but still fun). You've rediscovered the
rule-of-thumb that says that short beams usually fail design review in
stress, long beams in maximum deflection. Finally, the load is concentrated
at the ends and the center over a small length of the beam. You need to
treat that short length of the beam as a column and calculate the vertical
load that it will support without crumpling, and maybe add spreader plates
to apply the load over more of the beam area. All this is just for the
beam. Right now your beam will handle about 4800 lbs, so you need to either
limit the load or go to a bigger beam. Then you have to check the vertical
columns (your 2.5" square tube) for tensile strength, and the holes in the
columns and the pins that will support the beam for shear strength. Then
..., :-)

--
Regards,
Carl Ijames

When you do any structural calculation, you need to check all the
different modes that it can fail. For a simple beam you need to
do deflection, max stress, and end shear. For economy of time,
you pick the one that is most likely to fail for a given material
and do that first. If it works out, do the others. Steel is quite
stiff so you normally do the stress calc first, then the shear,
deflection last. If the first two work ok, the deflection is
usually ok also. Wood is the opposite: deflection is usually the
problem, especially with longer spans. I always do the deflection
calc first on wood unless I don't care if the structure bounces.

You asked why if the deflection is ok, why can't you use it?
Because you exceeded the elastic limit and the beam does not
spring back to the original shape. In fact, it will just keep
bending if you keep pushing the jack handle.

The beam you selected is an 'S' shape which is getting to be
relatively uncommon. Makes me think that this is a salvaged beam
rather than a new "A36" beam. As such, you have no idea what kind
of steel is in it. If it is salvage (or worse, salvage from a
fire), I would not assume it was up in the 45kpsi yield area.

Sounds like you are making a standard shop press. You will want
the extra tonnage so keep the 20 ton jack. Beam strength is a
1/x^2 function so shortening it up a little makes a big
difference. Take it down to 30" and you increase the theoretical
strength by 60%. The posts look fine for tension loads but you
will need to use large pins or reinforce the holes where the pins
go through to keep them from deforming. 1/2" pins will shear,
5/8" will want to bend.

BTW: 'bend like a pretzil' was overstating it. It would just bend
until you quit honking on the handle!

Cheers.

Steve wrote:

I posted a question about an I beam with a Moment of Inertia of 26.49.
It is a mild steel I-Beam and is 3.5" wide and 6" high. The beam
length is 38" and will have a 40,000 lb load placed in the middle of
the beam with a .5" spreader plate
where the jack will attach (This is going to be the top beam for a
hydraulic press). The deflection at load is .05". The beam will be
supported by two 60" posts that are 2.5"x2.5"x.25" thick square posts.
Someone states that this would "bend like a pretzel" under full load.
If the deflection is only .05" at maximum load how will the beam "bend
like a pretzel"? I know that the elastic yield strength of mild steel
is around 36,000 psi but I have read that most steel these days is
around the 45,000 psi. The same poster stated that the max stress will
be 43.6kpsi. I could truss the beam or could make the beam shorter in
length or could lower the hydraulic jack to a 10 ton model. I tried to
post this using the original post but was unsuccessful. Sorry for top
posting. Thanks, Steve.

18,000psi is a rather low value. Typical building codes allow
22kpsi for standard beams or 24kpsi for certain 'compact' shapes
in A36 steel. These numbers have a reasonable safety allowance
built in. If you actually put them in a tensile tester, you will
get higher values. So it boils down to what kind of safety factor
do you want to build in.

For new, prime A36 steel I would expect to see numbers up in the
40-45kpsi yield/55k tensile as measured in a good testing
machine. A36 is a pretty loose spec plus many steel yards call
any standard shape beam "A36" to distinguish it from HSLA or
other stock. Once you get to seconds, old, used, fire recovery,
or other salvage, all bets are off. And it usually goes down,not
up.

The OP is talking about an odd 'S' shape which is not a common
size in new prime stock. So I would guess old material and fudge
the number downward accordingly. On the other hand, if he bends
the top beam on a shop press, he sees it instantly and can just
quit pumping the handle, little harm done.

Increasing the depth of the beam has the same proportional effect
as shortening it. In this case, I suspect the OP has the chunk of
steel and wants to use it.

Cheers.

Old Nick wrote:

On 8 Aug 2003 23:41:30 -0700, (Steve) wrote
something
......and in reply I say!:

I think I can explain.

I am not sure where you get the figure of 45000 for mild steel.

I only play around with this stuff from time to time, for projects
that I build. But I had a bit of learning to do.

Standard steel (which you have to assume this is) can go as low as
15000 psi stress max. beam.exe quotes 18000 psi.

Beam depth is far more effective than shortening for this. I tried
your beam even at 24", and beam.exe says it will still fail at 40000
lb if standard steel (18000 psi from beam.exe). But an 8x4 beam will
_just_ sneak in at full length. A 10" by 5 " beam is well under.

I can see a danger of the beam still failing if any side force is
applied while it's under extreme strain. I have a huge press frame
that I picked up at an auction. It has twin beams, side by side, I
think for this reason.

I posted a question about an I beam with a Moment of Inertia of 26.49.
It is a mild steel I-Beam and is 3.5" wide and 6" high. The beam
length is 38" and will have a 40,000 lb load placed in the middle of
the beam with a .5" spreader plate
where the jack will attach (This is going to be the top beam for a
hydraulic press). The deflection at load is .05". The beam will be
supported by two 60" posts that are 2.5"x2.5"x.25" thick square posts.
Someone states that this would "bend like a pretzel" under full load.
If the deflection is only .05" at maximum load how will the beam "bend
like a pretzel"? I know that the elastic yield strength of mild steel
is around 36,000 psi but I have read that most steel these days is
around the 45,000 psi. The same poster stated that the max stress will
be 43.6kpsi. I could truss the beam or could make the beam shorter in
length or could lower the hydraulic jack to a 10 ton model. I tried to
post this using the original post but was unsuccessful. Sorry for top
posting. Thanks, Steve.

************************************************** ****************************************
I could never _see_ myself as anything!

Nick White --- HEAD:Hertz Music
Please remove ns from my header address to reply via email
!!
")
_/ )
( )
_//- \__/

On Sun, 10 Aug 2003 10:08:02 -0500, Roy Jenson
wrote something
.......and in reply I say!:

I am but an egg in this matter, seriously, and just plug in the
figures and see what happens.

But aren't we supposed to be looking at shear (or at least the Yield)
stress max, not tensile? I assumed that because the yield limit is
quite clearly shown as 18000 in the little programme I use. I looked
up some other data, and I have one steel that has a Yield of 300 MPa
and a Tensile of 450. The Yield is what I assumed was used to measure
bending strength.

Again plugging in the figures, if I halve the length, with the same
force and same beam, I halve the stress. If I double the depth,
leaving the length at its original value, the stress is one quarter,
from what I am seeing. So as far as I can see, depth (square) is more
useful than shortening (linear?).

I am not just being argumentative here, I thought I had this sorted
out, and I see a couple of other posts to the same thread with
different opinions.

18,000psi is a rather low value. Typical building codes allow
22kpsi for standard beams or 24kpsi for certain 'compact' shapes
in A36 steel. These numbers have a reasonable safety allowance
built in. If you actually put them in a tensile tester, you will
get higher values. So it boils down to what kind of safety factor
do you want to build in.

For new, prime A36 steel I would expect to see numbers up in the
40-45kpsi yield/55k tensile as measured in a good testing
machine. A36 is a pretty loose spec plus many steel yards call
any standard shape beam "A36" to distinguish it from HSLA or
other stock. Once you get to seconds, old, used, fire recovery,
or other salvage, all bets are off. And it usually goes down,not
up.

The OP is talking about an odd 'S' shape which is not a common
size in new prime stock. So I would guess old material and fudge
the number downward accordingly. On the other hand, if he bends
the top beam on a shop press, he sees it instantly and can just
quit pumping the handle, little harm done.

Increasing the depth of the beam has the same proportional effect
as shortening it. In this case, I suspect the OP has the chunk of
steel and wants to use it.

Cheers.

Old Nick wrote:

On 8 Aug 2003 23:41:30 -0700, (Steve) wrote
something
......and in reply I say!:

I think I can explain.

I am not sure where you get the figure of 45000 for mild steel.

I only play around with this stuff from time to time, for projects
that I build. But I had a bit of learning to do.

Standard steel (which you have to assume this is) can go as low as
15000 psi stress max. beam.exe quotes 18000 psi.

Beam depth is far more effective than shortening for this. I tried
your beam even at 24", and beam.exe says it will still fail at 40000
lb if standard steel (18000 psi from beam.exe). But an 8x4 beam will
_just_ sneak in at full length. A 10" by 5 " beam is well under.

I can see a danger of the beam still failing if any side force is
applied while it's under extreme strain. I have a huge press frame
that I picked up at an auction. It has twin beams, side by side, I
think for this reason.

I posted a question about an I beam with a Moment of Inertia of 26.49.
It is a mild steel I-Beam and is 3.5" wide and 6" high. The beam
length is 38" and will have a 40,000 lb load placed in the middle of
the beam with a .5" spreader plate
where the jack will attach (This is going to be the top beam for a
hydraulic press). The deflection at load is .05". The beam will be
supported by two 60" posts that are 2.5"x2.5"x.25" thick square posts.
Someone states that this would "bend like a pretzel" under full load.
If the deflection is only .05" at maximum load how will the beam "bend
like a pretzel"? I know that the elastic yield strength of mild steel
is around 36,000 psi but I have read that most steel these days is
around the 45,000 psi. The same poster stated that the max stress will
be 43.6kpsi. I could truss the beam or could make the beam shorter in
length or could lower the hydraulic jack to a 10 ton model. I tried to
post this using the original post but was unsuccessful. Sorry for top
posting. Thanks, Steve.

************************************************** ****************************************
I could never _see_ myself as anything!

Nick White --- HEAD:Hertz Music
Please remove ns from my header address to reply via email
!!
")
_/ )
( )
_//- \__/

************************************************** ****************************************
I could never _see_ myself as anything!

Nick White --- HEAD:Hertz Music
Please remove ns from my header address to reply via email
!!
")
_/ )
( )
_//- \__/

But aren't we supposed to be looking at shear (or at least the Yield)
stress max, not tensile? I assumed that because the yield limit is
quite clearly shown as 18000 in the little programme I use. I looked

Use the yield point, that is where it will bend and not spring
back. Actual values to use vary by grade of steel and may have
some safety factor inserted. 18kpsi is a resonably conservative
number for building construction. For new A36 steel I would
expect to see something like 40kpsi (or more) when I put a test
strip cut from the outer web in the pull test machine at school.
We usually see something like 50kpsi for hot rolled strap in the
same test.

Your program shows A36 at 18kpsi. My program comes preloaded at
24kpsi. Actual test is in the 40kpsi range. Yours has a bigger
safety factor loaded in.

As a practical matter, you won't see these low numbers in A36. we
needed some special steel for a very agressive forming operation.
Paying (lots of)extra money we could get some fully annealed,
fine grain stock that had a yield under 30kpsi but elongation of
around 40% in two inches without breaking.


up some other data, and I have one steel that has a Yield of 300 MPa
and a Tensile of 450. The Yield is what I assumed was used to measure
bending strength.

Again plugging in the figures, if I halve the length, with the same
force and same beam, I halve the stress. If I double the depth,
leaving the length at its original value, the stress is one quarter,
from what I am seeing. So as far as I can see, depth (square) is more
useful than shortening (linear?).

Deflection is a cubed function of the depth, stress is a squared
function, length is a LINEAR function. I didn't have enough
coffee on the first response, I used the linear loading function,
not the point load function. Which is why good engineers double
check their calculations the next day. My oops.


I am not just being argumentative here, I thought I had this sorted
out, and I see a couple of other posts to the same thread with
different opinions.

Cheers.

18,000psi is a rather low value. Typical building codes allow
22kpsi for standard beams or 24kpsi for certain 'compact' shapes
in A36 steel. These numbers have a reasonable safety allowance
built in. If you actually put them in a tensile tester, you will
get higher values. So it boils down to what kind of safety factor
do you want to build in.

For new, prime A36 steel I would expect to see numbers up in the
40-45kpsi yield/55k tensile as measured in a good testing
machine. A36 is a pretty loose spec plus many steel yards call
any standard shape beam "A36" to distinguish it from HSLA or
other stock. Once you get to seconds, old, used, fire recovery,
or other salvage, all bets are off. And it usually goes down,not
up.

The OP is talking about an odd 'S' shape which is not a common
size in new prime stock. So I would guess old material and fudge
the number downward accordingly. On the other hand, if he bends
the top beam on a shop press, he sees it instantly and can just
quit pumping the handle, little harm done.

Increasing the depth of the beam has the same proportional effect
as shortening it. In this case, I suspect the OP has the chunk of
steel and wants to use it.

Cheers.

Old Nick wrote:

On 8 Aug 2003 23:41:30 -0700, (Steve) wrote
something
......and in reply I say!:

I think I can explain.

I am not sure where you get the figure of 45000 for mild steel.

I only play around with this stuff from time to time, for projects
that I build. But I had a bit of learning to do.

Standard steel (which you have to assume this is) can go as low as
15000 psi stress max. beam.exe quotes 18000 psi.

Beam depth is far more effective than shortening for this. I tried
your beam even at 24", and beam.exe says it will still fail at 40000
lb if standard steel (18000 psi from beam.exe). But an 8x4 beam will
_just_ sneak in at full length. A 10" by 5 " beam is well under.

I can see a danger of the beam still failing if any side force is
applied while it's under extreme strain. I have a huge press frame
that I picked up at an auction. It has twin beams, side by side, I
think for this reason.

I posted a question about an I beam with a Moment of Inertia of 26.49.
It is a mild steel I-Beam and is 3.5" wide and 6" high. The beam
length is 38" and will have a 40,000 lb load placed in the middle of
the beam with a .5" spreader plate
where the jack will attach (This is going to be the top beam for a
hydraulic press). The deflection at load is .05". The beam will be
supported by two 60" posts that are 2.5"x2.5"x.25" thick square posts.
Someone states that this would "bend like a pretzel" under full load.
If the deflection is only .05" at maximum load how will the beam "bend
like a pretzel"? I know that the elastic yield strength of mild steel
is around 36,000 psi but I have read that most steel these days is
around the 45,000 psi. The same poster stated that the max stress will
be 43.6kpsi. I could truss the beam or could make the beam shorter in
length or could lower the hydraulic jack to a 10 ton model. I tried to
post this using the original post but was unsuccessful. Sorry for top
posting. Thanks, Steve.

************************************************** ****************************************
I could never _see_ myself as anything!

Nick White --- HEAD:Hertz Music
Please remove ns from my header address to reply via email
!!
")
_/ )
( )
_//- \__/

************************************************** ****************************************
I could never _see_ myself as anything!

Nick White --- HEAD:Hertz Music
Please remove ns from my header address to reply via email
!!
")
_/ )
( )
_//- \__/